centralizer Centralizer 2002-04-15 16:15:27 2004-05-24 16:03:39 Definition \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} Let $G$ be a group. The {\em centralizer} of an element $a \in G$ is defined to be the set $$C(a) = \{x \in G \mid xa = ax\}$$ Observe that, by definition, $e \in C(a)$, and that if $x, y \in C(a)$, then $xy^{-1}a = xy^{-1}a(yy^{-1})=xy^{-1}yay^{-1}=xay^{-1} = axy^{-1}$, so that $xy^{-1} \in C(a)$. Thus $C(a)$ is a subgroup of $G$. For $a \neq e$, the subgroup is non-trivial, containing at least $\{e, a\}$. To illustrate an application of this concept we prove the following lemma. {\bf Lemma:}\\ There exists a bijection between the right cosets of $C(a)$ and the conjugates of $a$. {\bf Proof:}\\ If $x,y \in G$ are in the same right coset, then $y = cx$ for some $c \in C(a)$. Thus $y^{-1}ay = x^{-1}c^{-1}acx = x^{-1}c^{-1}cax = x^{-1}ax$. Conversely, if $y^{-1}ay = x^{-1}ax$ then $xy^{-1}a = axy^{-1}$ and $xy^{-1} \in C(a)$ giving $x,y$ are in the same right coset. Let $[a]$ denote the conjugacy class of $a$. It follows that $|[a]| = [G : C(a)]$ and $|[a]| \mid |G|$.\\ We remark that $a \in Z(G) \iff C(a) = G \iff |[a]| = 1$, where $Z(G)$ denotes the center of $G$. Now let $G$ be a $p$-group, i.e. a finite group of order $p^n$, where $p$ is a prime and $n$ is a positive integer. Let $z = |Z(G)|$. Summing over elements in distinct conjugacy classes, we have $p^n = \sum{|[a]|} = z + \sum_{a \notin Z(G)}{|[a]|}$ since the center consists precisely of the conjugacy classes of cardinality $1$. But $|[a]| \mid p^n$, so $p \mid z$. However, $Z(G)$ is certainly non-empty, so we conclude that every $p$-group has a non-trivial center. The groups $C(gag^{-1})$ and $C(a)$, for any $g$, are isomorphic.